A triangle has sides with lengths 4.15, 7.092 and 12.45. What is the perimeter of the triangle?​

Answer: 24

Step-by-step explanation: negative multiples by a negative is a positive

The opposite of -(-24) is -24.

A number is additive inverse of a given number when sum of the two number is zero and a number is a multiplicative inverse of a given number when product of two number is 1.

According to the given question statement we have to determine what is opposite of -(-24).

We know opposite of two numbers sums to zero.

Given -(-24).              ( negative signs will distribute and it'll become 24).

Let the opposite number be x.

∴ 24 + x = 0

x = -24.

So the opposite of -(-24) is - 24.

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Because of the 45 45 90 theorem since the hypotenuse is 9√2 x and y are both 9

Answer: The unit price is $6.59 per pair of socks

Step-by-step explanation: 26.36. divided by 4 = 6.59

Answer:

3

Step-by-step explanation:

cause its a pretty simple question

Explanation:

The slope of the line is found by the method: rise/run

You calculate how much the slope rises(goes up) and divide it by run(goes horizontal).

One more thing to note is that: the slope is negative BECAUSE ITS DECREASING.

Your answer:   -3/2

Answer:

8.no answer A. x=8

Step-by-step explanation:

10.no answer B. x= -3

Answer:

8. (A) x = 8 , 10. (C.) x = 3

Step-by-step explanation:

Number 8 you could do 75 - 35 = 40

so what can you multiply 5 by (x) to get 40? you would multiply 5 by 8 or (x) and get 40

then add 35 + 40 to get 75 degrees which is the parallel angle to the other 75 degrees.

Number 10 you just want to get them to the same number.

so you would do ( 97 - 3 ) to get the ANSWER 87.

so x = 3

The answer is A

Hope this helps!

Answer:

2.94 in

Step-by-step explanation:

Answer:

61.7

Step-by-step explanation:

Answer:

See explanation

Step-by-step explanation:

Solving (a):

[tex]a_n = 6a_{n-1}[/tex] where [tex]a_0 = 2[/tex]

n = 1

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_1 = 6a_{1-1}[/tex]

[tex]a_1 = 6a_{0}[/tex]

Substitute 2 for [tex]a_0[/tex]

[tex]a_1= 6 * 2[/tex]

[tex]a_1= 12[/tex]

n = 2

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_2 = 6a_{2-1}[/tex]

[tex]a_2 = 6a_{1}[/tex]

Substitute 12 for [tex]a_1[/tex]

[tex]a_2= 6 * 12[/tex]

[tex]a_2= 72[/tex]

n = 3

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_3 = 6a_{3-1}[/tex]

[tex]a_3 = 6a_2[/tex]

Substitute 72 for [tex]a_2[/tex]

[tex]a_3= 6 * 72[/tex]

[tex]a_3= 432[/tex]

n = 4

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_4 = 6a_{4-1}[/tex]

[tex]a_4 = 6a_{3}[/tex]

Substitute 432 for [tex]a_3[/tex]

[tex]a_4 = 6 * 432[/tex]

[tex]a_4 = 2592[/tex]

n = 5

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_5 = 6a_{5-1}[/tex]

[tex]a_5 = 6a_4[/tex]

Substitute 2592 for [tex]a_4[/tex]

[tex]a_5 = 6 * 2592[/tex]

[tex]a_5 = 15552[/tex]

n = 6

[tex]a_n = 6a_{n-1}[/tex]

[tex]a_6 = 6a_{6-1}[/tex]

[tex]a_6 = 6a_{5}[/tex]

Substitute 15552 for [tex]a_5[/tex]

[tex]a_6 = 6 * 15552[/tex]

[tex]a_6 = 93312[/tex]

[tex]a_1= 12[/tex]   [tex]a_2= 72[/tex]   [tex]a_3= 432[/tex]   [tex]a_4 = 2592[/tex]   [tex]a_5 = 15552[/tex]   [tex]a_6 = 93312[/tex]

Solving (b):

[tex]a_n = a^2_{n - 1[/tex] where [tex]a_1 = 2[/tex]

We have

[tex]a_1 = 2[/tex] which serves as the first term

n =2

[tex]a_n = a^2_{n - 1[/tex]

[tex]a_2 = a^2_{2-1}[/tex]

[tex]a_2 = a^2_{1}[/tex]

Substitute 2 for [tex]a_1[/tex]

[tex]a_2 = 2^2[/tex]

[tex]a_2 = 4[/tex]

n = 3

[tex]a_3 = a^2_{3-1}[/tex]

[tex]a_3 = a^2_{2}[/tex]

[tex]a_3 = 4^2[/tex]

[tex]a_3 = 16[/tex]

n = 4

[tex]a_4 = a^2_{4-1}[/tex]

[tex]a_4 = a^2_3[/tex]

[tex]a_4 = 16^2[/tex]

[tex]a_4 = 256[/tex]

n =5

[tex]a_5 = a^2_{5-1[/tex]

[tex]a_5 = a^2_4[/tex]

[tex]a_5 = 256^2[/tex]

[tex]a_5 = 65536[/tex]

n = 6

[tex]a_6 = a^2_{6-1[/tex]

[tex]a_6 = a^2_{5[/tex]

[tex]a_6 = 65536^2[/tex]

[tex]a_6 = 4294967296[/tex]

[tex]a_1 = 2[/tex]   [tex]a_2 = 4[/tex]   [tex]a_3 = 16[/tex]   [tex]a_4 = 256[/tex]   [tex]a_5 = 65536[/tex]   [tex]a_6 = 4294967296[/tex]

Solving (c):

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] [tex]a_0=1[/tex]  ; ;  [tex]a_1=2[/tex]

[tex]a_1=2[/tex] ---- First term

n = 2

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_2=a_{2-1}+3a_{2-2}[/tex]

[tex]a_2=a_1+3a_0[/tex]

Substitute values for a1 and a0

[tex]a_2=2+3 * 1[/tex]

[tex]a_2=2+3[/tex]

[tex]a_2=5[/tex]

n = 3

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_3=a_{3-1}+3a_{3-2}[/tex]

[tex]a_3=a_{2}+3a_{1}[/tex]

[tex]a_3=5+3 * 2[/tex]

[tex]a_3=5+6[/tex]

[tex]a_3=11[/tex]

n = 4

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_4=a_{4-1}+3a_{4-2}[/tex]

[tex]a_4=a_{3}+3a_{2}[/tex]

[tex]a_4=11+3 * 5[/tex]

[tex]a_4=11+15[/tex]

[tex]a_4=26[/tex]

n = 5

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_5=a_{5-1}+3a_{5-2};[/tex]

[tex]a_5=a_{4}+3a_3[/tex]

[tex]a_5=26+3 * 11[/tex]

[tex]a_5=26+33[/tex]

[tex]a_5=59[/tex]

n = 6

[tex]a_n=a_{n-1}+3a_{n-2};[/tex] becomes

[tex]a_6=a_{6-1}+3a_{6-2}[/tex]

[tex]a_6=a_{5}+3a_4[/tex]

[tex]a_6=59+3*26[/tex]

[tex]a_6=59+78[/tex]

[tex]a_6=137[/tex]

[tex]a_1=2[/tex]   [tex]a_2=5[/tex]   [tex]a_3=11[/tex]   [tex]a_4=26[/tex]   [tex]a_5=59[/tex]   [tex]a_6=137[/tex]

Solving (d):

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex];     [tex]a_0=1[/tex]; [tex]a_1=1[/tex]

[tex]a_1=1[/tex] --- First term

n = 2

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_2=2 * a_{2-1}+2^2a_{2-2}[/tex]

[tex]a_2=2 * a_1+4*a_0[/tex]

[tex]a_2=2 * 1+4*1[/tex]

[tex]a_2=2 +4[/tex]

[tex]a_2=6[/tex]

n = 3

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_3=3 * a_{3-1}+3^2 * a_{3-2}[/tex]

[tex]a_3=3 * a_{2}+9 * a_{1}[/tex]

[tex]a_3=3 * 6+9 * 1[/tex]

[tex]a_3=18+9[/tex]

[tex]a_3=27[/tex]

n = 4

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_4=4*a_{4-1}+4^2*a_{4-2}[/tex]

[tex]a_4=4*a_{3}+16*a_{2}[/tex]

[tex]a_4=4*27+16*6[/tex]

[tex]a_4=204[/tex]

n = 5

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_5=5 * a_{5-1}+5^2 * a_{5-2}[/tex]

[tex]a_5=5 * a_{4}+25 * a_{3}[/tex]

[tex]a_5=5 * 204+25 *27[/tex]

[tex]a_5=1695[/tex]

n = 6

[tex]a_n=na_{n-1}+n^2a_{n-2}[/tex] becomes

[tex]a_6=6 * a_{6-1}+6^2*a_{6-2}[/tex]

[tex]a_6=6 * a_{5}+36*a_{4}[/tex]

[tex]a_6=6 * 1695+36*204[/tex]

[tex]a_6=17514[/tex]

[tex]a_1=1[/tex]   [tex]a_2=6[/tex]   [tex]a_3=27[/tex]   [tex]a_4=204[/tex]   [tex]a_5=1695[/tex]   [tex]a_6=17514[/tex]

Solving (e):

[tex]a_n= a_{n-1}+a_{n-3};\ a_0=1; a_1=2; a_2=0[/tex]

First term: [tex]a_1=2[/tex]

Second Term: [tex]a_2=0[/tex]

n = 3

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_3= a_{3-1}+a_{3-3}[/tex]

[tex]a_3= a_{2}+a_0[/tex]

[tex]a_3= 0+1[/tex]

[tex]a_3= 1[/tex]

n = 4

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_4= a_{4-1}+a_{4-3}[/tex]

[tex]a_4= a_{3}+a_{1}[/tex]

[tex]a_4= 1+2[/tex]

[tex]a_4=3[/tex]

n = 5

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_5= a_{5-1}+a_{5-3}[/tex]

[tex]a_5= a_{4}+a_{2}[/tex]

[tex]a_5= 3 + 0[/tex]

[tex]a_5= 3[/tex]

n = 6

[tex]a_n= a_{n-1}+a_{n-3}[/tex] becomes

[tex]a_6= a_{6-1}+a_{6-3}[/tex]

[tex]a_6= a_{5}+a_{3}[/tex]

[tex]a_6= 3 + 1[/tex]

[tex]a_6= 4[/tex]

[tex]a_1=2[/tex]   [tex]a_2=0[/tex]   [tex]a_3= 1[/tex]   [tex]a_4=3[/tex]   [tex]a_5= 3[/tex]   [tex]a_6= 4[/tex]

Answer:

It is $15.00

Step-by-step explanation:

It is B because A is not even money, C is fifteen cents which is not correct because the question is asking for 75% of $20.00, and finally D is incorrect because it is more than the amount of money asked in the question.

Answer:

Option C. 2(x + 8)

Step-by-step explanation:

To know which option is the correct answer, let us simplify each expression given in the options to know which will not result to 2x² – 16. This is illustrated below:

For Option A:

2x² + 4x – 4x – 16

= 2x² – 16

For Option B:

2(x² – 8)

Clear bracket

2x² – 16

For Option C:

2(x + 8)

Clear bracket

2x + 8

For Option D:

(4x² – 32) /2

4x²/2 – 32/2

2x² – 16

From the above illustration, option C did not result to 2x² – 16.

Therefore, option C is the answer to the question.

Answer:

Option (4) and Option (6)

Step-by-step explanation:

By the zero product property,

If (x - a)(x - b) = 0

Then, (x - a) = 0 ⇒ x = a

(x - b) = 0 ⇒ x = b

By following the zero product property,

From the given equation,

3(x - 4)(2x - 3) = 0

x - 4 = 0 ⇒ x = 4

2x - 3 = 0 ⇒ x = [tex]\frac{3}{2}[/tex]

Therefore, options (4) and (6) will be the correct options.

Answer:

4 and 6 :)

Step-by-step explanation:

edge

Answer:

She will pay $2.

Step-by-step explanation:

Answer:

Standard Deviation and Below are, in fact, the right answer.

Step-by-step explanation:

Edge. :)

Answer:

The person above is right, they deserve brainliest.

-standard deviations

-below

ED2021

Answer:

1/26

Step-by-step explanation:

There are two sets colors; red and black.

13 ranks, each rank has 2 black cards and 2 red

but there are only 2 black eights

2/52

Answer:

Real

Step-by-step explanation:

i^6-122-7i^2

(i^2)^3-122-7i^2

(-1)^3-122-7×(-1)

-1-122+7

-123+7

-116

So the final answer is -116

Answer:

a. Yes

Step-by-step explanation:

The x intercept is (9/2,0) or (4.5,0). The y intercept is (0,-1):)

Answer:  

470

Step-by-step explanation:

Replace x with 10:  470

5(10)(10 - 3( = 500 - 30 = 470

470

Answer:

p=2 (l+w)

Step-by-step explanation:

Answer:

Yes, because the plot shows no apparent pattern.

Step-by-step explanation:

Yes, because the plot is a residual plot from a regression of a variable with the independent variable x.

Therefore, the correct option is 1.)Yes, because the plot shows no apparent pattern.

To learn more about regression refer to:

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Answer:

2) 9/4

Step-by-step explanation:

12/5 in decimal form is 2.4. As a mixed number, 12/5 is 2 2/5. And, 12/5 is just a reduced form of 36/15. So, the only answer it can be is answer 2: 9/4

Answer:

Answer is 79 and 28

Step-by-step explanation:

Let two numbers be x and y.

x + y = 107. --> 1

x - y = 51 -->2

solving above equations,

2x = 158

x =79

y = 28

Answer:

diameter ÷ 2 = 8 ÷ 2 = 4

Step-by-step explanation:

Answer:

9. -3/4

10. -5/1

11. UNDEFINED

12. 2/3

Answer:

9. -3/4

10. -5

11. undefined

12. 2/3

Step-by-step explanation:

Answer:

Mrs Barnes has the highest

Answer:

Answers below

Step-by-step explanation:

Equation

Let's call:

p = number of pennies Janesa has in her pocket

n = number of nickels Janesa has in her pocket

She has 24 coins in total, thus:

[tex]p + n=24\qquad\qquad [1][/tex]

Since each penny is worth $0.01 and each nickel is worth $0.05, and Janesa has $1 in total:

[tex]0.01p+0.05n=1[/tex]

Multiplying by 100:

[tex]p+5n=100\qquad\qquad [2][/tex]

A. Completing the equation:

[tex]5n+p=100[/tex]

B.

From [1]:

[tex]p=24-n[/tex]

Substituting into [2]:

[tex]24-n+5n=100[/tex]

Simplifying:

[tex]24+4n=100[/tex]

Rearranging:

[tex]4n=100-24=76[/tex]

[tex]n=76 / 4= 19[/tex]

[tex]p=24-19=5[/tex]

B. There are 5 pennies in Janesa's pocket

C. Since each penny is worth $0.01, the total value of pennies is 5*$0.01=$0.05

The total value of the pennies is $0.05

D. The total value of the nickels is 19*$0.05 = $0.95

E. Jackson should say she has 19 nickels and 5 pennies

janesa has 24 coins in her pocket worth exactly $1.00. she tells her younger brother jackson that he can have the coins if he correctly guesses how many of each coin she has. she gives him the hint that there are only nickels and pennies. complete the equation that represents the total value of the nickels and p pennies 5( blank ) + p = 100 The type of coins that Janessa has = nickels and pennies.

Janessa has a total number of 24 coins

The sum of the nickels and pennies = $1.00

Let us represent the number of nickels as x

The number of pennies as y